A man finished a job in 5 days. On the first day, he finished 1/m of the job. On the second day, he finished 1/n of the job left. On the third day, he finished 1/m of the job left, and on the fourth day, 1/n of the job left. On the last day, he finished 1/4 of the remaining job. If and , find and .
Answers
Given : A man finished a job in 5 days.
On the first day, he finished 1/m of the job.
On second day, he finished 1/n of the job left.
On the third day, he finished 1/m of the job left, and on the fourth day 1/n of the job left.
On the last day, 1/4 of the job was left to be done.
m,n ∈ N and n>m,
To find : m and n.
Solution:
Work done on 1st day = 1/m
Left = 1 - 1/m = (m - 1)/m
Work done on second day = (1/n) ( (m - 1)/m)
Work Left = (m - 1)/m - (1/n) ( (m - 1)/m)
= ( (m - 1)/m)( (n - 1)/n)
= (m - 1)(n - 1) / mn
Similarly after 4 days work left = {(m - 1)(n - 1) / mn }² = 1/4
=> (m - 1)(n - 1) / mn = 1/2
=> mn - m - n + 1 = mn/2
=> 2mn - 2m - 2n + 2 = mn
=> mn + 2 = 2m + 2n
2m + 2n = 2 + mn
m + n = 1 + mn/2
as 1/2 work is completed in 2 days Hence m must be greater than 2
m = 3 =>
3 + n = 1 + 3n/2
=> n/2 = 2
=> n = 4
m = 4 =>
4 + n = 1 + 2n
=> n = 3
m = 3 and n = 4 satisfy this as n > m
m =3 and n = 4
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